Calculate the pH value of 0.15 M solution of acetic acid.
`K_(CH_3COOH)=1.8xx10^(-5)`

Answer

To calculate the pH of a 0.15 M solution of acetic acid (CH₃COOH) with a dissociation constant \( K_a = 1.8 \times 10^{-5} \), we can follow these steps:

### Step 1: Write the dissociation equation
Acetic acid dissociates in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Set up the equilibrium expression
The equilibrium expression for the dissociation of acetic acid is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]
Where:
- \([\text{CH}_3\text{COO}^-]\) and \([\text{H}^+]\) are the concentrations of acetate ions and hydrogen ions at equilibrium.
- \([\text{CH}_3\text{COOH}]\) is the concentration of acetic acid at equilibrium.

### Step 3: Define initial concentrations and changes
Let the initial concentration of acetic acid be \( C = 0.15 \, \text{M} \). At equilibrium, if \( \alpha \) is the degree of dissociation, we have:
- \([\text{CH}_3\text{COOH}] = C(1 - \alpha) = 0.15(1 - \alpha)\)
- \([\text{CH}_3\text{COO}^-] = [\text{H}^+] = C\alpha = 0.15\alpha\)

### Step 4: Substitute into the equilibrium expression
Substituting these values into the \( K_a \) expression gives:
\[ K_a = \frac{(0.15\alpha)(0.15\alpha)}{0.15(1 - \alpha)} = \frac{0.0225\alpha^2}{0.15(1 - \alpha)} \]
Setting this equal to \( K_a \):
\[ 1.8 \times 10^{-5} = \frac{0.0225\alpha^2}{0.15(1 - \alpha)} \]

### Step 5: Simplify the equation
Multiply both sides by \( 0.15(1 - \alpha) \):
\[ 1.8 \times 10^{-5} \times 0.15(1 - \alpha) = 0.0225\alpha^2 \]
This simplifies to:
\[ 2.7 \times 10^{-6}(1 - \alpha) = 0.0225\alpha^2 \]

### Step 6: Assume \( \alpha \) is small
Since acetic acid is a weak acid, we can assume \( \alpha \) is small, thus \( 1 - \alpha \approx 1 \):
\[ 2.7 \times 10^{-6} \approx 0.0225\alpha^2 \]

### Step 7: Solve for \( \alpha \)
Rearranging gives:
\[ \alpha^2 = \frac{2.7 \times 10^{-6}}{0.0225} \approx 1.2 \times 10^{-4} \]
Taking the square root:
\[ \alpha \approx \sqrt{1.2 \times 10^{-4}} \approx 0.011 \]

### Step 8: Calculate \([\text{H}^+]\)
Now, we can find the concentration of hydrogen ions:
\[ [\text{H}^+] = C\alpha = 0.15 \times 0.011 \approx 0.00165 \, \text{M} \]

### Step 9: Calculate pH
Finally, we can calculate the pH:
\[ \text{pH} = -\log[\text{H}^+] = -\log(0.00165) \approx 2.78 \]

### Final Answer
The pH of the 0.15 M solution of acetic acid is approximately **2.78**.

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